A current i is uniformly distributed over the cross section of a long hollow cylinderical wire of inner radius `R_1` and outer radius `R_2`. Magnetic field B varies with distance r form the axis of the cylinder is

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Answer

To find the magnetic field \( B \) at a distance \( r \) from the axis of a long hollow cylindrical wire with inner radius \( R_1 \) and outer radius \( R_2 \), we can use Ampere's Law, which states:

\[ \oint B \cdot dl = \mu_0 I_{\text{inside}} \]

where \( I_{\text{inside}} \) is the current enclosed by the Amperian loop.

### Step 1: Analyze the regions

1. **Region 1: \( r < R_1 \)** (inside the hollow part)
- There is no current enclosed.
- By Ampere's Law:
\[ \oint B \cdot dl = 0 \implies B = 0 \]

2. **Region 2: \( R_1 < r < R_2 \)** (within the hollow cylinder)
- Here, we need to find the current enclosed by a circular Amperian loop of radius \( r \).
- The current density \( J \) is given by:
\[ J = \frac{I}{\pi(R_2^2 - R_1^2)} \]
- The area of the annular region from \( R_1 \) to \( r \) is:
\[ A = \pi(r^2 - R_1^2) \]
- The current enclosed by the Amperian loop is:
\[ I_{\text{inside}} = J \cdot A = \frac{I}{\pi(R_2^2 - R_1^2)} \cdot \pi(r^2 - R_1^2) = \frac{I(r^2 - R_1^2)}{R_2^2 - R_1^2} \]
- By applying Ampere's Law:
\[ B \cdot (2\pi r) = \mu_0 I_{\text{inside}} \implies B \cdot (2\pi r) = \mu_0 \frac{I(r^2 - R_1^2)}{R_2^2 - R_1^2} \]
- Thus,
\[ B = \frac{\mu_0 I (r^2 - R_1^2)}{2\pi r (R_2^2 - R_1^2)} \]

3. **Region 3: \( r \geq R_2 \)** (outside the hollow cylinder)
- The total current \( I \) is enclosed.
- By Ampere's Law:
\[ B \cdot (2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r} \]

### Summary of Results

- For \( r < R_1 \):
\[ B = 0 \]
- For \( R_1 < r < R_2 \):
\[ B = \frac{\mu_0 I (r^2 - R_1^2)}{2\pi r (R_2^2 - R_1^2)} \]
- For \( r \geq R_2 \):
\[ B = \frac{\mu_0 I}{2\pi r} \]